Cartesian product of two $G_\delta$ sets is measurable This was written in stein's real analysis, integratetion text (p85, prop 3.6)

Clearly, $G=G_1 \times G_2$ is measurable in $\mathbb{R}^{d_1} \times \mathbb{R}^{d_2}$, where $G_j \subseteq \mathbb{R}^{d_j}$ are of type $G_{\delta}$ (countable intersection of open sets.)

How is this true? 
My argument is as follows, 
$G_1 = \bigcap O_i $, $G_2 = \bigcap O'_j$, so 
$$G_1 \times G_2 = \bigcap O_i \times \bigcap O'_j= \bigcap O_i \times O'_i$$
which is a $G_{\delta}$ set in $\mathbb{R}^d$ ($d=d_1+d_2$) hence measurable. 
It seems weird to me that the second equality holds, but I argue as follows: 
\begin{align*}
(x,y) \in \bigcap_{i=1}^{\infty} O_i \times O'_i & \Leftrightarrow x \in \bigcap_{i=1}^{\infty} O_i, y \in \bigcap_{j=1}^{\infty} O'_j \\ 
& \Leftrightarrow (x,y) \in \bigcap_{i=1}^{\infty} O_i \times \bigcap_{j=1}^{\infty} O'_j
\end{align*}
 A: Your proof is probably fine, but the notation is sloppy. I'm answering here because I have some anxiety about your use of $i$ for every index, particularly when you're trying to understand the detail of what's happening. So, let's let $A$ and $B$ be two countable index sets and let $G_1 = \bigcap\limits_{\alpha \in A} O_\alpha$ and $G_2 = \bigcap\limits_{\beta \in B} O'_{\beta}$. Then, what you're trying to show is the equality
$$
\left[\bigcap\limits_{\alpha \in A} O_{\alpha}\right] \times \left[ \bigcap_{\beta \in B} O'_{\beta} \right] = \bigcap_{(\alpha,\beta) \in A\times B}O_{\alpha} \times O_{\beta}
$$
Expanding on your proof just a bit,
$$
(x,y) \in \left[\bigcap\limits_{\alpha \in A} O_{\alpha}\right] \times \left[ \bigcap_{\beta \in B} O'_{\beta} \right] \\
\iff\\
x \in \bigcap\limits_{\alpha \in A} O_{\alpha} \quad \text{and} \quad y \in  \bigcap_{\beta \in B} O'_{\beta} \\
\iff\\
(x,y) \in O_{\alpha}\times O_{\beta} ~\text{ for all } (\alpha, \beta) \in A \times B \\
\iff\\
(x,y) \in \bigcap_{(\alpha,\beta) \in A\times B}O_{\alpha} \times O_{\beta}
$$
